Training Heat Exchangers — LMTD Method Fouling and Design Considerations
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Fouling and Design Considerations

24 min Heat Exchangers — LMTD Method

Fouling — the accumulation of deposits on heat-transfer surfaces — increases thermal resistance and reduces exchanger performance over time. The fouling factor Rf is added to the overall resistance: 1/Udirty = 1/Uclean + Rf. Designers must account for fouling when specifying surface area, cleaning schedules, and materials of construction. Ignoring fouling leads to undersized exchangers that fail to meet process requirements as the equipment ages.

Fouling and Design Considerations

Real heat exchangers degrade over time as deposits build up on surfaces. Fouling reduces $U$, increases pressure drop, and raises operating costs. Engineers must account for fouling at the design stage.

Fouling Factor

$$\frac{1}{U_{\text{dirty}}} = \frac{1}{U_{\text{clean}}} + R''_{f,i} + R''_{f,o}$$

$R''_f$ = fouling resistance (m²·K/W). Added to both inner and outer surfaces as appropriate.

Typical Fouling Resistances

Distilled water: $0.0001$ • City water (below 50°C): $0.0002$ • City water (above 50°C): $0.0004$ • River water: $0.0003$–$0.001$ • Seawater: $0.0001$–$0.0003$ • Light oil: $0.0002$ • Heavy fuel oil: $0.0009$ • Steam (non-oil-bearing): $0.0001$

Units: m²·K/W

Example 1 — Effect of Fouling

A water-to-water HX has $U_{\text{clean}} = 1500$ W/(m²·K). Both sides foul with $R''_f = 0.0003$ m²·K/W. Find $U_{\text{dirty}}$ and the percentage reduction.

  1. $\frac{1}{U_{\text{dirty}}} = \frac{1}{1500} + 0.0003 + 0.0003 = 0.000667 + 0.0006 = 0.001267$
  2. $U_{\text{dirty}} = 789$ W/(m²·K)
  3. Reduction: $(1500 - 789)/1500 = 47\%$
  4. Fouling nearly halved the effective $U$!
Example 2 — Overdesign Factor

A clean HX needs $A = 8$ m². With fouling, $U$ drops by 30%. How much area is needed?

  1. If $U_{\text{dirty}} = 0.7 U_{\text{clean}}$, then to maintain the same $\dot{Q}$:
  2. $A_{\text{dirty}} = A_{\text{clean}} / 0.7 = 8/0.7 = 11.4$ m²
  3. Overdesign factor = $11.4/8 = 1.43$ (43% excess area).
  4. This is typical for industrial practice.
Example 3 — Pressure Drop Consideration

A double-pipe HX has $\dot{Q} = 50$ kW, $U = 400$, $\Delta T_{\text{lm}} = 30$°C. Current tube velocity is 1.5 m/s. If we double the number of tubes (keeping same total flow), what happens?

  1. $A_{\text{required}} = 50{,}000/(400 \times 30) = 4.17$ m² (unchanged)
  2. Doubling tubes: velocity halves (0.75 m/s), so $h$ decreases (turbulent $h \propto V^{0.8}$): $h_{\text{new}}/h_{\text{old}} = 0.5^{0.8} = 0.574$.
  3. Lower $h$ means lower $U$, requiring more area.
  4. But pressure drop $\propto V^2$ decreases 4×, reducing pump power.
  5. This is the classic heat transfer vs. pressure drop trade-off.

Practice Problems

1. $U_{\text{clean}} = 2000$. Inner fouling $R''_f = 0.0002$, outer $R''_f = 0.0005$. Find $U_{\text{dirty}}$.
2. $\dot{Q} = 80$ kW, $U_{\text{dirty}} = 500$, $\Delta T_{\text{lm}} = 20$. Area needed?
3. If fouling reduces $U$ by 40%, by what factor must $A$ increase to maintain $\dot{Q}$?
4. Heavy oil ($R''_f = 0.0009$) vs. distilled water ($R''_f = 0.0001$). Which needs more overdesign?
5. Why is fouling worse at higher temperatures?
6. An HX is overdesigned by 25%. It can tolerate fouling that reduces $U$ by what percentage?
Show Answer Key

1. $1/U = 1/2000 + 0.0002 + 0.0005 = 0.0005 + 0.0007 = 0.0012$. $U = 833$ W/(m²·K)

2. $A = 80{,}000/(500 \times 20) = 8$ m²

3. $A_{\text{new}}/A_{\text{old}} = 1/0.6 = 1.667$ (67% more area)

4. Heavy oil — 9× higher fouling resistance.

5. Higher temps accelerate scaling, corrosion, biological growth, and chemical reactions that form deposits.

6. $\dot{Q} = UA\Delta T_{\text{lm}}$. With 25% more area: $1.25 U_{\text{dirty}} A_{\text{needed}} = U_{\text{clean}} A_{\text{needed}}$. So $U_{\text{dirty}} = U_{\text{clean}}/1.25 = 0.80U$. Tolerates 20% reduction in $U$.

🚿 Fouling Impact Calculator
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Shell-and-Tube and Cross-Flow Correction Factors Practice Test — Heat Exchangers (LMTD)